Question

Take the Hadamard product for $\zeta(s)$:

$$\displaystyle \zeta(s) = \pi^{\frac{s}{2}} \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)}{2(s-1)\Gamma(1+\frac{s}{2})}$$

and reshuffle it into:

$$\displaystyle \dfrac{\prod_\rho \left(1- \frac{s}{\rho} \right) \left(1- \frac{s}{1-\rho} \right)} {\zeta(s)} = \dfrac{2(s-1)\Gamma(1+\frac{s}{2})}{ \pi^{\frac{s}{2}}}$$

After experimenting with some values for $s$, I found that f.i. $\zeta(3)$ can be simply written as:

$$\zeta(3) = \prod_\rho \left(1- \frac{2}{\rho} \right) \left(1- \frac{2}{1-\rho} \right)\left(1- \frac{3}{\rho} \right) \left(1- \frac{3}{1-\rho} \right)$$

and believe this can be generalized into ($k= 1,2,3...$):

$$\displaystyle \zeta(2k+1) = a[2k+1] \prod_\rho \left(1- \frac{2k}{\rho} \right) \left(1- \frac{2k}{1-\rho} \right)\left(1- \frac{2k+1}{\rho} \right) \left(1- \frac{2k+1}{1-\rho} \right)$$

with $a[3]=1, a[5]=\frac12, a[7]=\frac15, a[9]=\frac{5}{84}, \dots$

1) Is this a known result? Could the $a[2k+1]$ sequence be derived from the Bernoulli numbers?

2) Could it be proven that when solving $\rho$ for one factor of the infinite product, i.e. only equate the 4 subfactors to $\zeta(2k+1)$, that the complex roots always must have $\Re(s)=\frac12$ ?

Answer 1

(1) Yes: multiplying your Hadamard products for $\zeta(2k+1)$ and $\zeta(2k)$ and then dividing by the known formula $\zeta(2k) = |B_{2k}|(2\pi)^{2k}/2(2k)!$, we obtain $$ \zeta(2k+1) = \frac1{(2k-1)2k(2k+1)|B_{2k}|} \prod_\rho \bigg( 1-\frac{2k}\rho \bigg) \bigg( 1-\frac{2k}{1-\rho} \bigg) \bigg( 1-\frac{2k+1}\rho \bigg) \bigg( 1-\frac{2k+1}{1-\rho} \bigg). $$

(2) What?